Recent content by seand

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    Solving 2tv' - v = 0: Is it Separable?

    well I think I want to switch things around so v and t are separated 1/v dv = 1/2t dt integrating both sides ln(v) = 1/2 ln(t)+C e^x both sides: v = k*sqrt(t) Which looks right! Is that how I was meant to do it? If so, thanks - I got stalled before when I got the ln() on both sides.
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    Solving 2tv' - v = 0: Is it Separable?

    How do you find a solution for: 2tv' - v = 0 The text says it's separable but I'm not seeing it. I'm just learning so extra details are appreciated. Thanks. (this should have been posted in the homework section - but I can't seem to move it there, sorry)
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    Undergrad What is the point in calculating orthonormal bases for R3

    I'm just learning linear algebra, where we are taking a set of basis vectors and then using gram schmidt to convert them to an orthonormal basis. So I know I can do it. I know orthonormal is great. But as I'm modifying the basis anyway from the original, why don't I just forget the original...
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    Is this a simple circle, with radius 1/2?

    Completing the square worked perfectly. Thank you.
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    Is this a simple circle, with radius 1/2?

    I'm trying to understand this expression: y=x^2+y^2 Using wolfram alpha it tells me this is a circle of radius 1/2 But to me I'm unfamiliar with this way of setting up a circle. Specifically the y on both sides of the equation is troubling me. How would you interpret it? Would you...
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    Undergrad Chain rule notation - can Leibniz form be made explicit?

    Thank you Hurkyl, Landau, mathman and Tiny Tim for the replies and especially the suggestions. It's interesting to know that I could express the chain rule the way I did, but that there are other, more normal ways to do it.
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    Undergrad Chain rule notation - can Leibniz form be made explicit?

    Hi there, I'm a new user to the forums (and Calculus) and I 'm hoping you can give me your opinion on my chain rule form below. When learning the chain rule, I was taught two forms. This form: \frac{d}{dx}f(g(x))=f'(g(x))g'(x) As well as the Leibniz form...